The three lines `4x - 7y + 10 , x + y = 5 and 7x+ 4y = 15 ` form the sides of a triangle Then the point (1,2) is its

Options

A. centroid
B. incentre
C. orthocentre
D. none of these

Answer

To determine the status of the point (1, 2) with respect to the triangle formed by the lines \(4x - 7y + 10 = 0\), \(x + y = 5\), and \(7x + 4y = 15\), we will follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are:
1. \(4x - 7y + 10 = 0\) (Equation 1)
2. \(x + y - 5 = 0\) (Equation 2)
3. \(7x + 4y - 15 = 0\) (Equation 3)

### Step 2: Find the intersection points of the lines
To find the vertices of the triangle, we need to calculate the intersection points of the lines.

#### Intersection of Line 1 and Line 2
From Equation 1:
\[ 4x - 7y + 10 = 0 \Rightarrow 4x - 7y = -10 \]
From Equation 2:
\[ x + y = 5 \Rightarrow y = 5 - x \]

Substituting \(y\) in Equation 1:
\[ 4x - 7(5 - x) = -10 \]
\[ 4x - 35 + 7x = -10 \]
\[ 11x - 35 = -10 \]
\[ 11x = 25 \]
\[ x = \frac{25}{11} \]

Now substituting \(x\) back to find \(y\):
\[ y = 5 - \frac{25}{11} = \frac{55}{11} - \frac{25}{11} = \frac{30}{11} \]

So, the intersection point \(A\) is:
\[ A\left(\frac{25}{11}, \frac{30}{11}\right) \]

#### Intersection of Line 1 and Line 3
From Equation 1:
\[ 4x - 7y = -10 \]
From Equation 3:
\[ 7x + 4y = 15 \]

We can multiply Equation 1 by 4 and Equation 3 by 7 to eliminate \(y\):
\[ 16x - 28y = -40 \quad (Equation 4) \]
\[ 49x + 28y = 105 \quad (Equation 5) \]

Adding Equation 4 and Equation 5:
\[ 16x - 28y + 49x + 28y = -40 + 105 \]
\[ 65x = 65 \]
\[ x = 1 \]

Substituting \(x = 1\) back into Equation 1:
\[ 4(1) - 7y + 10 = 0 \]
\[ 4 - 7y + 10 = 0 \]
\[ 14 - 7y = 0 \]
\[ 7y = 14 \]
\[ y = 2 \]

So, the intersection point \(B\) is:
\[ B(1, 2) \]

#### Intersection of Line 2 and Line 3
From Equation 2:
\[ x + y = 5 \Rightarrow y = 5 - x \]
From Equation 3:
\[ 7x + 4y = 15 \]

Substituting \(y\) in Equation 3:
\[ 7x + 4(5 - x) = 15 \]
\[ 7x + 20 - 4x = 15 \]
\[ 3x + 20 = 15 \]
\[ 3x = -5 \]
\[ x = -\frac{5}{3} \]

Now substituting \(x\) back to find \(y\):
\[ y = 5 - \left(-\frac{5}{3}\right) = 5 + \frac{5}{3} = \frac{15}{3} + \frac{5}{3} = \frac{20}{3} \]

So, the intersection point \(C\) is:
\[ C\left(-\frac{5}{3}, \frac{20}{3}\right) \]

### Step 3: Determine the nature of the point (1, 2)
Now we have the vertices of the triangle:
- \(A\left(\frac{25}{11}, \frac{30}{11}\right)\)
- \(B(1, 2)\)
- \(C\left(-\frac{5}{3}, \frac{20}{3}\right)\)

Since point \(B(1, 2)\) is one of the vertices of the triangle formed by the three lines, it is a vertex of the triangle.

### Conclusion
The point (1, 2) is a vertex of the triangle formed by the lines.